More information is needed to solve the problem.
You question implies simple subtraction is needed:
28-8=20
We know that
A line tangent to a circle is perpendicular to the radius to the point of tangency.
so
<span>Line ef is </span>perpendicular to the segment gh
hence
Triangle FHG is a right triangle
We have to prove that rectangles are parallelograms with congruent Diagonals.
Solution:
1. ∠R=∠E=∠C=∠T=90°
2. ER= CT, EC ║RT
3. Diagonals E T and C R are drawn.
4. Shows Quadrilateral R E CT is a Rectangle.→→[Because if in a Quadrilateral One pair of Opposite sides are equal and parallel and each of the interior angle is right angle than it is a Rectangle.]
5. Quadrilateral RECT is a Parallelogram.→→[If in a Quadrilateral one pair of opposite sides are equal and parallel then it is a Parallelogram]
6. In Δ ERT and Δ CTR
(a) ER= CT→→[Opposite sides of parallelogram]
(b) ∠R + ∠T= 90° + 90°=180°→→→Because RECT is a rectangle, so ∠R=∠T=90°]
(c) Side TR is Common.
So, Δ ERT ≅ Δ CTR→→[SAS]
Diagonal ET= Diagonal CR →→→[CPCTC]
In step 6, while proving Δ E RT ≅ Δ CTR, we have used
(b) ∠R + ∠T= 90° + 90°=180°→→→Because RECT is a rectangle, so ∠R=∠T=90°]
Here we have used ,Option (D) : Same-Side Interior Angles Theorem, which states that Sum of interior angles on same side of Transversal is supplementary.
Let's take a peek of each term, to be specific, the exponent of the variables only, to see what degree each term is
so...which term has the highest degree? well, that one, is the degree of the polynomial